Kronecker Delta
Definition
$\delta_{ij}$ defined as follows is called the Kronecker delta.
$$ \delta_{ij} := \begin{cases} 1,&i=j \\ 0, & i\ne j \end{cases} $$
Explanation
The Kronecker delta is used in a great many places, and its main role is to pick out only the one you want among all the components (elements, possibilities, etc.). Physics students usually first meet it as a way of expressing the dot product. Since it may not be clear what this means, let us understand it by looking at the example below.
Example
First, suppose two vectors $\mathbf{A}=(A_{1}, A_{2}, A_{3})$ and $\mathbf{B}=(B_{1}, B_{2}, B_{3})$ are given. Then the dot product of the two vectors is as follows.
$$ \mathbf{A} \cdot \mathbf{B} = A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3} $$
Expressing this with the summation symbol $\sum$ gives the following.
$$ \mathbf{A} \cdot \mathbf{B} = A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3} = \sum \limits_{i=1}^{3}A_{i}B_{i} $$
Then one can see through the following that the expression above and $\sum \limits_{i=1}^{3}\sum \limits_{j=1}^{3}\delta_{ij}A_{i}B_{j}$ are the same expression.
$$ \begin{align*} \sum _{i=1}^{3}\sum _{j=1}^{3}\delta_{ij}A_{i}B_{j} &= \delta_{11}A_{1}B_{1} + \delta_{12}A_{1}B_{2} + \delta_{13}A_{1}B_{3} \\ & \quad+ \delta_{21}A_{2}B_{1} + \delta_{22}A_{2}B_{2} + \delta_{23}A_{2}B_{3} \\ & \quad+ \delta_{31}A_{3}B_{1} + \delta_{32}A_{3}B_{2} + \delta_{33}A_{3}B_{3} \\ &= 1\cdot A_{1}B_{1} + 0 \cdot A_{1}B_{2} + 0\cdot A_{1}B_{3} \\ & \quad+ 0\cdot A_{2}B_{1} + 1\cdot A_{2}B_{2} + 0\cdot A_{2}B_{3} \\ & \quad+ 0\cdot A_{3}B_{1} + 0\cdot A_{3}B_{2} + 1\cdot A_{3}B_{3} \\ &= A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3} \\ &= \sum \limits_{i=1}^{3}A_{i}B_{i} \\ &= \mathbf{A} \cdot \mathbf{B} \end{align*} $$
Applying to the result above the Einstein notation, which omits $\sum$ when the same index appears two or more times on one side, gives the following.
$$ \delta_{ij}A_{i}B_{j} = \mathbf{A} \cdot \mathbf{B} $$
So one may accept that $\delta_{ij}A_{i}B_{j}$ and $\mathbf{A} \cdot \mathbf{B}$ are the same, yet still not understand why such an expression is used. Since the example above is a very simple formula, its usefulness does not stand out, but once you compute dot products of countless vectors along with cross products, gradients, divergences, curls, Laplacians, and so on in electromagnetism, you will come to appreciate its convenience. If you are a second-year undergraduate, you will naturally come to know that convenience, so there is no need to force yourself to accept it right now.
Also, since a value exists only when both subscripts are the same, if two or more Kronecker deltas are multiplied as below, then naturally a value exists only when all the indices are the same.
$$ \delta_{ij}\delta_{jk} $$
In a case like the above, a nonzero value exists only when $i=j=k$. The Kronecker delta is also an example of a $2$nd-order tensor.
One can also understand this by representing it as a matrix.
Formulas
For $i,j \in \left\{ 1,2,3 \right\}$ the following formulas hold. Do not forget that $\sum$ is omitted on a side where one index appears two or more times.
(a) $\delta_{ii} = 3$
(b) $\delta_{ij}\delta_{lj} = \delta_{ij}\delta_{jl} = \delta_{il}$
(c) $\delta_{ii}\delta_{jj} = 9$
(d) $\delta_{ii}\delta_{jj} = 6 \quad (i \ne j)$
(e) $\epsilon_{ijk}\delta_{k\ell} = \epsilon_{ij\ell}$
Here $\epsilon_{ijk}$ is the Levi-Civita symbol.
Proof
(a)
By Einstein notation the following holds.
$$ \delta_{ii} = \sum \limits_{i=1}^{3} \delta_{ii} = \delta_{11}+\delta_{22}+\delta_{33}=3 $$
■
(b)
By Einstein notation the following holds.
$$
\delta_{ij}\delta_{jl}=\sum\limits_{j=1}^{3}\delta_{ij}\delta_{jl}=\delta_{i1}\delta_{1l}+\delta_{i2}\delta_{2l}+\delta_{i3}\delta_{3l}
$$
Now let us think about the cases in which the value above is not $0$. There are the following three cases.
$$ i=l=1 \quad \text{and} \quad i=l=2 \quad \text{and} \quad i=l=3 $$
In the first case, the following holds.
$$ \delta_{i1}\delta_{1l} = 1 \quad \text{and} \quad \delta_{i2}\delta_{2l}=\delta_{i3}\delta_{3l} = 0 \\ \implies \delta_{ij}\delta_{jl} = \delta_{i1}\delta_{1l}+\delta_{i2}\delta_{2l}+\delta_{i3}\delta_{3l} = 1 $$
In the second case, the following holds.
$$ \delta_{i2}\delta_{2l} = 1 \quad \text{and} \quad \delta_{i1}\delta_{1l}=\delta_{i3}\delta_{3l} = 0 \\ \implies \delta_{ij}\delta_{jl} = \delta_{i1}\delta_{1l}+\delta_{i2}\delta_{2l}+\delta_{i3}\delta_{3l} = 1 $$
In the third case, the following holds.
$$ \delta_{i3}\delta_{3l} = 1 \quad \text{and} \quad \delta_{i1}\delta_{1l}=\delta_{i2}\delta_{2l} = 0 \\ \implies \delta_{ij}\delta_{jl} = \delta_{i1}\delta_{1l}+\delta_{i2}\delta_{2l}+\delta_{i3}\delta_{3l} = 1 $$
Therefore $\delta_{ij}\delta_{jl}$ has the value $1$ only when $i=l$, and in all the other cases its value is $0$, so we obtain the following result.
$$ \delta_{ij}\delta_{jl} = \delta_{il} $$
■
(c)
Since $\sum$ is omitted by Einstein notation, we have the following.
$$ \begin{align*} \delta_{ii}\delta_{jj} &= \sum\limits_{i=1}^{3}\sum\limits_{j=1}^3{\delta_{ii}\delta_{jj}} \\ &= \sum\limits_{i=1}^{3}{\delta_{ii} \sum\limits_{j=1}^3\delta_{jj}} \\ &= 3\cdot 3 \\ &= 9 \end{align*} $$
The third equality holds by (a) .
■
(d)
Since $\sum$ is omitted by Einstein notation, we have the following.
$$ \begin{align*} \delta_{ii}\delta_{jj} &= \sum\limits_{i=1}^{3}\sum\limits_{\substack{j=1 \\ j\ne i}}^{3}{\delta_{ii}\delta_{jj}} \\ &= \delta_{11}\delta_{22} +\delta_{11}\delta_{33} +\delta_{22}\delta_{11} +\delta_{22}\delta_{33}+\delta_{33}\delta_{11}+\delta_{33}\delta_{22} \\ &= 6 \end{align*} $$
■
(e)
Since $\sum$ over $k$ is omitted by Einstein notation, we have the following.
$$ \epsilon_{ijk}\delta_{k\ell} = \sum\limits_{k=1}^{3} \epsilon_{ijk}\delta_{k\ell} = \epsilon_{ij1}\delta_{1\ell} + \epsilon_{ij2}\delta_{2\ell} + \epsilon_{ij3}\delta_{3\ell} $$
Since the Kronecker delta has the value $1$ only when the two indices are the same and $0$ in the other cases, only the one term with $k = \ell$ remains among the three terms above. Writing this out according to the values $\ell$ can take gives the following.
$$ \begin{align*} \ell = 1 &: \quad \epsilon_{ij1} \cdot 1 + \epsilon_{ij2} \cdot 0 + \epsilon_{ij3} \cdot 0 = \epsilon_{ij1} \\ \ell = 2 &: \quad \epsilon_{ij1} \cdot 0 + \epsilon_{ij2} \cdot 1 + \epsilon_{ij3} \cdot 0 = \epsilon_{ij2} \\ \ell = 3 &: \quad \epsilon_{ij1} \cdot 0 + \epsilon_{ij2} \cdot 0 + \epsilon_{ij3} \cdot 1 = \epsilon_{ij3} \end{align*} $$
In all three cases the remaining value is $\epsilon_{ij\ell}$, so we obtain the following result.
$$ \epsilon_{ijk}\delta_{k\ell} = \epsilon_{ij\ell} $$
The same argument goes through even if an object with some index stands in place of the Levi-Civita symbol, so contraction with the delta ultimately plays the role of changing that index from $k$ to $\ell$.
■